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Write $P=(P_{ij}{)}_{i,j=1}^3$, where $P_{ij}:{\mathbb{C}}^{m_j}\to {\mathbb{C}}^{m_i}$. Block multiplication makes the commuting condition equivalent toThus every block is an intertwiner. For $i\ne j$, the irreducible representations are inequivalent, so the Schur lemma gives $P_{ij}=0$. On each diagonal block, the same Schur lemma gives $P_{ii}={\lambda }_i{I}_{m_i}$. ThereforeConversely, every displayed matrix commutes with $R(g)$, proving both necessity and sufficiency. This commutant of a semisimple representation has complex dimension three. The parameters may vanish: $P$ was not required to be invertible.

Now $m_1=m_2=n$ and the first two matrix representations agree in the chosen bases. Each of $P_{11},P_{12},P_{21},P_{22}$ is therefore a commuting endomorphism of that common irreducible representation, so each is scalar by the Schur lemma. All blocks connecting either of these copies with the third vanish, since the third irreducible representation is inequivalent. HenceEvery such matrix commutes with $R(g)$. In particular the off-diagonal scalars $b,c$ can mix the two identical copies. The commutant of a semisimple representation is $M_2(\mathbb{C})\oplus \mathbb{C}$, of complex dimension five. If the first two representations were merely equivalent, rather than literally equal, the mixed blocks would instead be scalar multiples of a fixed invertible intertwiner and its inverse.

All three vector spaces now have the same dimension $n$. Each of the nine blocks of $P$ commutes with the common irreducible representation, so the Schur lemma givesConversely $R(g)=I_3\otimes R_1(g)$, so the Kronecker product identity immediately verifies commutation for every $A$. The commutant of a semisimple representation acts arbitrarily on the three-dimensional multiplicity vector space while acting as a scalar on each irreducible factor. Its complex dimension is nine.

Partition $M$ into the same five-by-five blocks as $R$. The equation for block $(i,j)$ is $M_{ij}{R}_j(g)=R_i(g)M_{ij}$, so each block is an intertwiner. By Schur's lemma, blocks between the two inequivalent irreducible representations vanish, whereas blocks between identical copies are arbitrary scalar multiples of the relevant identity matrix. Thus the commutant of a semisimple representation isEquivalently, $M=(A\otimes I_{n_1})\oplus (B\otimes I_{n_2})$, where $A\in M_3(\mathbb{C})$ and $B\in M_2(\mathbb{C})$. Substitution proves every such matrix commutes with every $R(g)$. There is no invertibility requirement on $M$, and its solution space has complex dimension $3^2+2^2=13$, even if $n_1=n_2$.